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Is the function $ f(x,y)=xy/ (x^{2}+y^{2})$ where $f(0,0)$ is defined to be $0$ continuous? I don't think it is and I am trying to either show this by the definition or by showing that maybe a close set in $\mathbb{R}$ has an inverse set that is not closed in $ \mathbb{R} ^{2}$. I tried the point $0$ but this is open in $\mathbb{R}$ Any hints or ideas? Thanks !

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up vote 11 down vote accepted

It is not continuous at $(0,0)$. Because $f(t,t)=1/2$ but $f(t,0)=0$, so if we approach to the origin along the line $y=x$ then $f(x,y)\to 1/2$ but if we approach to the origin along the x-axis then $f(x,y)\to 0$.

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Nice, thank you. – Jmaff Oct 8 '13 at 2:01

If you use polar coordinates

$$ f(r\cos(\theta), r\sin(\theta))= {\sin(\theta)\cos(\theta)}=\frac{\sin(2\theta)}{2}, $$

then, you can see that the limit has infinite number of values depending on the choice of theta which implies the limit does not exist. A related problem.

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Well, If it were continuous then for ALL $\epsilon >0$, there exist $\delta>0$ such that if $||x-a||<\delta$ then $||f(x)-f(a)||<\epsilon$. So it is enough to show one counter example. For a specific epsilon, say $\epsilon =\frac{1}{4}$, then clearly $$||(\frac{\delta}{2}, \frac{\delta}{2})-(0,0)|| = ||(\frac{\delta}{2},\frac{\delta}{2})|| = \frac{\delta}{\sqrt{2}}<\delta$$

But $$||\frac{\frac{\delta}{2}\frac{\delta}{2}}{\frac{\delta^2}{4}{}\frac{\delta^2}{4}}- 0||=\frac{1}{2}>\frac{1}{4} = \epsilon$$ There you go you have a counter example here where we picked a specific epsilon and could not conclude $||f(x)-f(a)||<\epsilon$

The above answers are correct but you should be able to prove this using delta epsilon definition like I did. Note that the answers above say you approach the origin along y=x suggests the delta I pick for the proof. You see I picked $(\frac{\delta}{2}, \frac{\delta}{2})$ in the form of $(x,x)$ for my $(x,y)$ since I'm approaching the origin along y=x.

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Consider an approach along the line $y = x$; then

$$f(x, y) = \frac{x^2}{x^2 + x^2} = \frac{1}{2}$$

for all $x \ne 0$. On the other hand, if we approach $(0, 0)$ along the line $y = 2x$,

$$f(x, y) = \frac{2x^2}{x^2 + 4x^2} = \frac{2}{5}$$

So there are two different paths toward the origin, each giving a different limit. Hence, the limit does not exist.

Or more simply, approach along $y = 0$.

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$f(x,y) = xy/(x^2+y^2)$ is continuous when $x \to 0$ and $y \to 0$.

A function is said to be contineous if limit exists and $$ \begin{array}{rcl} \lim_{\substack{x \to a\\ y \to b}} f(x,y) & = & \lim_{\substack{x \to b\\ y \to a}} f(x,y) \\ \lim_{\substack{x \to a\\ y \to b}} \frac{xy}{x^2+y^2} & = & \lim_{\substack{x \to b\\ y \to a}} \frac{xy}{x^2+y^2} \\ \lim_{\substack{x \to a\\ y \to b}} \frac{x0}{x^2+0} & = & \lim_{\substack{x \to b\\ y \to a}} \frac{0y}{0+y^2} \\ \lim_{\substack{x \to a\\ y \to b}} \frac{0}{x^2} & = & \lim_{\substack{0 \to b\\ y \to a}} \frac{0}{y^2} \\ 0 & = & 0 \end{array} $$

Note $0/x=0$.

The value of function is given which is zero. Hence limit of function is equal to value of function, so it is continuous.

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$f$ is cotinuous at $(a,b)$ if $\lim\limits_{(x,y)\to (a,b)}f(x,y)=f(a,b)$. This is a stronger condition than taking the limit with respect to $x$ then $y$, or vice versa. For example, the function in this question is not continuous. It takes on the value $\frac12$ arbitrarily close to $(0,0)$, but $f(0,0)=0$. – Jonas Meyer Oct 15 '14 at 0:44

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